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4 years ago

15

Calculate the amount of heat, in kilojoules, required to turn solid ammonia at -78 degrees Celsius to gaseous ammonia at standar

d temperature

1 answer:

zhuklara [117]4 years ago

3 0

Answer:

132

Explanation:

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When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a

Triss [41]

Answer:

26.85kJ/mol is the heat of dissolution of Cs₂O₄

Explanation:

The process of dissolution of Cs₂SO₄ in water occurs as follows:

Cs₂O₄(s) → 2Cs⁺(aq) + SO₄²⁻(aq) + ΔH

<em>Where ΔH is the change in heat per mole of </em>Cs₂O₄

The first we can see is ΔH > 0 because heat is absorbed (Temperature is decreasing) when the reaction occurs.

Now, the change in heat of reaction is:

q = Heat calorimeter + Heat solution

q = ΔT*1.65J/°C + S*m×ΔT

<em>Where q is the heat of reaction.</em>

<em>ΔT is change in temperature: 25.62°C - 22.57°C = 3.05°C</em>

<em>S is specific heat of solution = Specific heat water = 4.184J/g°C</em>

<em>m is mass of solution = 22.24g + 106.70g = 128.94g</em>

Replacing, heat of reaction is:

q = 3.05°C*1.65J/°C + 4.184J/g°C*128.94g×3.05°C

q = 1650J are absorbed when 22.24g of Cs₂O₄ reacts

Moles of 22.24g of Cs₂O₄ are - Molar mass: 361.87g/mol-:

22.24g * (1mol / 361.87g) = 0.06146 moles

That means, when 0.06146 moles of Cs₂O₄ react, the heat absorbed is 1650J. That means the heat absorbed per mole of Cs₂O₄ (Enthalpy of dissolution) is:

1650J / 0.06146 moles = 26847J / mol =

<h3>26.85kJ/mol is the heat of dissolution of Cs₂O₄</h3>

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6 0

3 years ago

Which of the following is NOT an example of a chemical change?

Fudgin [204]

Its c im pretty sure

8 0

3 years ago

Read 2 more answers

How many grams of Ca(OH)2are required to make 1.5 L of a 0.81 M solution?

FrozenT [24]

Answer:

Mass = 90.28 g

Explanation:

Given data:

Mass of Ca(OH)₂ = ?

Volume of solution= 1.5 L

Molarity of solution = 0.81 M

Solution:

First of all we will calculate number of moles.

Molarity = number of moles / volume in L

by putting values,

0.81 M = Number of moles / 1.5 L

Number of moles = 0.81 M × 1.5 L

Number of moles = 1.22 mol

Mass of Ca(OH)₂ in gram:

Mass = number of moles × molar mass

Mass = 1.22 mol × 74.09 g/mol

Mass = 90.28 g

5 0

3 years ago

Based on Reference Table F, which solution will contain the highest concentration of

wariber [46]

Iron (II) iodide solution will contain the highest concentration of iodide ions. Option D is the right answer.

Explanation:

Reference table F is used to check the solubility of the metal compounds formed. It tells if the compound is soluble or insoluble in water.

From the table, it is seen that Mercury, silver and lead form insoluble complex with iodide ions. This means that they have low concentration of dissolved iodide ions. These three metals lead, silver and mercury are given as exception in the table F.

Thus iron(II) iodide will have highest concentration of iodide ions when dissolved in water.

7 0

3 years ago

The Haber reaction for the manufacture of ammonia is: N2 + 3H2 → 2NH3 Without doing any experiments, which of the following can

Dimas [21]

Answer : The correct statement is, $\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }N_2=-\frac{d[N_2]}{dt}$

$\text{Rate of disappearance of }H_2=-\frac{1}{3}\frac{d[H_2]}{dt}$

$\text{Rate of formation of }NH_3=+\frac{1}{2}\frac{d[NH_3]}{dt}$

From this we conclude that,

$\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)$

Hence, the correct statement is, $\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)$

3 0

3 years ago